= Solution
Assume the usual nontrivial case $\chi(G)\ge2$ and $|H|\ge2$. On $(\chi(G)-1)(|H|-1)$ vertices partition into $\chi(G)-1$ equal classes of size $|H|-1$. Color all edges between distinct classes red and all edges within a class blue. The red <graph> is $(\chi(G)-1)$-colorable, so it contains no copy of $G$. Every blue <connected> component has only $|H|-1$ vertices, so <connectedness> prevents a blue copy of $H$. Therefore one more vertex is necessary:
$$
\boxed{R(G,H)\ge(\chi(G)-1)(|H|-1)+1}.
$$
If $G$ is edgeless or $H$ has one vertex, the displayed lower bound is just one and holds directly.
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